Question

Prove that the general equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ will represent two parallel straight lines if $h^2=ab$ and $b{g}^2=a{f}^2$. Also prove that the distance between them is $2\sqrt{\left\{\frac{g^2-ac}{a(a+b)}\right\}}$.
Also prove that $\frac{a}{h}=\frac{h}{b}=\frac{g}{f}$.

Answer 1

We can show that the general equation of second degree will represent a pair of parallel straight lines if $h^2=ab$ or $h=\sqrt{\left(ab\right)}$. In order to find the second condition, put the value of h in the equation.
$\therefore a{x}^2+2\sqrt{\left(ab\right)}xy+b{y}^2+2gx+2fy+x=0$
or $(\sqrt{\left(ax\right)}+\sqrt{\left(by\right)}{]}^2+2gx+2fy+c=0$
Now suppose that above is equal to
$[\sqrt{\left(ax\right)}+\sqrt{\left(by\right)}+p][\sqrt{\left(ax\right)}+\sqrt{\left(by\right)}+q]=0$
Comparing the coefficients,
$p\sqrt{a}+q\sqrt{a}=2g$ or $(p+q)=\frac{2g}{\sqrt{a}}$ .$\left(1\right)$
$p\sqrt{b}+q\sqrt{b}=2f$ or $p+q=\frac{2f}{\sqrt{b}}$ $\left(2\right)$
$pq=c$ ..$\left(3\right)$
Now from $\left(1\right)$ and $\left(2\right)$,
$\frac{2g}{\sqrt{a}}=\frac{2f}{\sqrt{b}}$ or $b{g}^2=a{f}^2$
Distance between the lines
$=\frac{0+0+p}{\sqrt{(a+b)}}-\frac{0-0+q}{\sqrt{(a+b)}}=\frac{p-q}{\sqrt{(a+b)}}$
$=\frac{\left\{\right(p+q{)}^2-4pq{\}}^{1/2}}{\sqrt{(a+b)}}={\left\{\frac{\frac{4g^2}{a}-4.c}{(a+b)}\right\}}^{1/2}$ [by $\left(1\right)$ and $\left(3\right)$]
$=2\sqrt{\left\{\frac{g^2-ac}{a(a+b)}\right\}}$
Now from above, we have $h^2=ab$ and $b{g}^2=a{f}^2$
$\frac{a}{h}=\frac{h}{b}=k$ say,
$\therefore a=hk$ and $b=\frac{h}{k}$;
$\therefore$ from the condition $b{g}^2=a{f}^2$,
$\frac{h}{k}{g}^2=hk.f^2$;
$\therefore \frac{g^2}{f^2}=k^2$ or $\frac{g}{f}=k$;
$\therefore \frac{a}{h}=\frac{h}{b}=\frac{g}{f}$ as each is equal to k.